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A certain disease has an incidence rate of 0.9%.
If the false negative rate is 4% and the false positive rate is 5%,
compute the probability that a person who tests positive actually has the disease.
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My previous solution was incorrect, as tutor @math_tutor2020 noticed it.
Therefore, I removed / (deleted / erased) that version and placed here a new one, corrected and fixed.
So now you see the corrected and fixed version. Thanks to tutor @math_tutor2020.
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This problem is about calculating the CONDITIONAL PROBABILITY.
They want you calculate this ratio .
Let X be the total population.
Then the number of those who has the disease is 0.009*X; the number of those who has no the disease is 0.991*X.
The number of those who has the disease and tests positive is 0.009*(1-0.04)X = 0.009*0.96*X (true test positive)
(to get this amount, I exclude from the sick persons, 0.009*X, the number of false negative 0.009*x*0.04.
The number of those who has no the disease and tests positive is 0.991*0.05*X (false test positive).
So the number of those who tests positive is the sum 0.009*0.96*X + 0.991*0.05*X.
They want you calculate the conditional probability
P = = = cancel X in the numerator and in the denominator =
= = = use your calculator = 0.1485 (rounded). ANSWER
ANSWER. This probability is 0.1485.
Solved.